Calculus integration of the Gaussian distrib. bell curve??

Question

$$y=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$$

What do you mean with "impossible integration"? The primitive cannot be expressed using elementary functions (unless you count $\operatorname{erf}$ among the elementary functions), but for example $\int_0^z e^{-x^2/2},dx$ can be computed well enough. — Daniel Fischer, Dec 09 '13 at 23:48
How would you compute that integral? With u/du substitution? I tried to set u=$\frac{-x^2}{2}$ but then $du=-x dx$ — JackOfAll, Dec 10 '13 at 15:15
I'm not an expert for the error function, so I don't know what the professionals use to compute it. For $\lvert z\rvert < 1$, you could just integrate the power series term by term, the series converges fast enough to give good approximations with not too much computation. For large $z$, you can compute $$\int_z^\infty \frac1x \left(xe^{-x^2/2}\right),dx,$$ integrate by parts and get a good approximation of the result with not much computation. For $z$ from approximately $1$ to well, whatever counts as "large", I don't know off-hand how you quickly get good approximations, but there are ways. — Daniel Fischer, Dec 10 '13 at 15:32

score 2 · Accepted Answer · answered Dec 09 '13 at 23:49

2

The (indefinite) integral is the error function (erf). It can be proved that it cannot be expressed in terms of elementary functions.

answered Dec 09 '13 at 23:49

Igor Rivin

Ok, so you can really just do the definite integral? – JackOfAll Dec 10 '13 at 15:09

1 Answers1